Exercise 1.3.26


Theorem: Let and be categories, functors and a natural transformation. is a natural isomorphism if and only if is an isomorphism (in ) for each .

Proof: Suppose is a natural isomorphism. That is to say, it is an isomorphism between and in the category . Then, we have a natural transformation so that and (the identity natural transformations on and , respectively). Recall is a collection of arrows in for each . And, by definition, they compose “component-wise”, so that for each . Then, clearly we have , but is, by definition, precisely the identity on – that is, . Simultaneously, we have . These identities directly state that (and ) are isomorphisms between and .

Suppose that, for each , is an isomorphism. Then, also for each we have an arrow with and . The identity natural transformation is defined so that for each . So we have for each , which means that . Similarly, we obtain , and (and ) are natural isomorphisms.